Elimination Algebra 1

Question

You are running a concession stand at a basketball game. You are selling hotdogs and sodas. Each hotdog

costs $3.00 and each soda costs $2.00. At the end of the night, you made a total of $208. You sold a

combined total of 87 hotdogs and sodas. Write a system of equations to represent this situation.

Answer 1

Because this is a system, you need two inequalities- one to represent the cost (because we are given total cost) and one to represent the number (because we are given the total number of hot dogs and sodas).

For the cost we know the total is 208 bought for 3 dollars per hot dog and 2 dollars per soda. So you need to make a variable for hot dogs and for soda. So I'm going to say h= the number of hot dogs and s = the number of sodas.

Your first equation would look like

3h + 2s = 208

(the total cost of the hot dogs + the total cost of the sodas = you total money made)

For the total number we just know that the number of hot dogs and sodas together is 87. And we said h=# of hot dogs and s=# of sodas. So the second equation is

h + s = 87

And I believe that's all you had to do for that problem since it just says to write your system and not solve.

Answer 2

We can let h be the number of hotdogs sold and s be the number of sodas sold

The first equation is easy. Since we sold a combined total of 87 hotdogs and sodas....

h + s = 87

Since hotdogs cost $3 and sodas cost $2 and the total money collected was $208....

$3h + $2s = $208 or

3h + 2s = 208

So your system of equations will be

h + s = 87

3h + 2s = 208

or if you like we can just use the normal x and y variables (x = number of hotdogs, y = number of sodas)

x + y = 87

3x + 2y = 208